Chapter 5: Gases

5.1 Early Experiments 5.2 The gas laws of Boyle, Charles, and Avogadro 5.3 The Ideal Gas Law 5.4 Gas Stoichiometry 5.5 Dalton’s Law of Partial Pressures 5.6 The Kinetic molecular Theory of Gases 5.7 Effusion and Diffusion 5.8 Collisions of Gas Particles with the Container Walls 5.9 Intermolecular Collisions

•  Helium He 4.0 •  Neon Ne 20.2 •  Argon Ar 39.9 •  Hydrogen H2 2.0 •  Nitrogen N2 28.0 •  Nitrogen Monoxide NO 30.0 •  Oxygen O2 32.0 •  Hydrogen Chloride HCL 36.5 •  Ozone O3 48.0 •  Ammonia NH3 17.0 •  Methane CH4 16.0

Substances that are Gases under Normal Conditions

Substance Formula M (g/mol)

Important Characteristics of Gases

1) Gases are highly compressible An external force compresses the gas sample and decreases its volume, removing the external force allows the gas volume to increase. 2) Gases are thermally expandable When a gas sample is heated, its volume increases, and when it is cooled its volume decreases. 3) Gases have low viscosity Gases flow much easier than liquids or solids. 4) Most Gases have low densities Gas densities are on the order of grams per liter whereas liquids and solids are grams per cubic cm, 1000 times greater. 5) Gases are infinitely miscible

Gases mix in any proportion such as in air, a mixture of many gases.

Pressure = force per unit area (Definition!)

Gas samples exert a force pushing against any surface they contact, due to the impact of molecules striking the surface. Per unit area, it’s the same for all surfaces in or walls containing the same gas sample, unless the gas is flowing somewhere collectively.

That pressure increases with molecular velocity and mass, and with the number of molecules per unit volume (number density).

A sheet suspended within a gas sample gets same force on both sides, so it does not move.

Average molecular velocity increases with temperature but decreases with molecular mass, in such a way that the mass effect cancels.

This leaves pressure nearly proportional to temperature and to number density, but independent of the molecular mass:

P = R(T)(n/V), where R is the gas constant, n = # moles = N/NA

Pressure = Force Area

Pressure of the Atmosphere

•  Called “Atmospheric pressure,” or the force exerted on earth’s surface by the gases in air

•  The force per unit area of these gases.

•  Can be measured using a barometer:

Pressure = Force Area

Common Units of Pressure

Unit Atmospheric Pressure Scientific Field Used

Pascal (Pa) = N/m2; 1.01325 x 105 Pa SI unit; physics, kilopascal (kPa) 101.325 kPa Chemistry

bar 1.01325 bar Meteorology, Chemistry

atmosphere (atm) 1 atm Chemistry

millimeters of mercury 760 mmHg Chemistry, medicine, (mmHg), also called 760 torr biology “torr”

pounds per square inch 14.7 lb/in2 Engineering (psi or lb/in2)

Converting Units of Pressure Problem: A chemist collects a sample of Carbon dioxide from the decomposition of Lime stone (CaCO3) in a closed-end manometer (i.e, vacuum is the reference pressure), the height of the mercury is 341.6 mm Hg. Calculate the CO2 pressure in torr, atmospheres, and kilopascals. Plan: The pressure is in mmHg, so we use the conversion factors from Table above to find the pressure in the other units. Solution:

PCO2 (torr) = 341.6 mm Hg x = 341.6 torr 1 torr 1 mm Hg

converting from mmHg to torr:

converting from torr to atm: PCO2( atm) = 341.6 torr x = 0.4495 atm 1 atm

760 torr converting from atm to kPa:

PCO2(kPa) = 0.4495 atm x = 45.54 kPa 101.325 kPa 1 atm

Boyle’s Law : P - V inverse proportional when n and T are constant in a gas sample

•  Pressure is inversely proportional to Volume •  P = or V = or PV=k •  Change of Conditions Problems if n and T are constant ! •  P1V1 = k P2V2 = k •  Then : P1V1 = P2V2

k V

k P

Figure5.4:Plo/ngBoyle’sdatafromTable5.1.

Applying Boyles Law to Gas Problems Problem: A gas sample at a pressure of 1.23 atm has a volume of 15.8 cm3, what will be the volume in L if the pressure is increased to 3.16 atm? Plan: We begin by converting the volume that is in cm3 to mL and then to liters, then we do the pressure change to obtain the final volume! Solution:

V1 (cm3)

V1 (ml)

V1 (L)

V2 (L)

1cm3 = 1 mL

1000mL = 1L

x P1/P2

P1 = 1.23 atm P2 = 3.16 atm V1 = 15.8 cm3 V2 = unknown T and n remain constant

V1 = 15.8 cm3 x x = 0.0158 L 1 mL 1 cm3

1 L 1000mL

V2 = V1 x = 0.0158 L x = 0.00615 L P1 P2

1.23 atm 3.16 atm

CharlesLaw‐V‐T‐proporDonal

•  Atconstantpressureandforafixedamount(#moles)ofgas:

VolumeisproporDonaltoTemperature: V=constantxAbsoluteTemperature

V=kT•  Changeofcondi@onsproblems:

•  SinceV=kTorV1/T1=V2/T2or:

T1

V1

T2 or =

V2

T1 = V1 x T 2

V2

Temperatures must be expressed in Kelvin!

Figure 5.7: Plots of V versus T (c) for several gases.

Figure 5.8: Plots of V versus T as in Fig. 5.7 except that here the Kelvin scale is used for temperature.

Charles Law Problem

•  A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant.

•  V1 = 3.20 L T1 = 125oC = 398 K •  V2 = 1.54 L T2 = ?

•  V1 / T1 = V2 / T2 So T2 =

Charles Law Problem

•  A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant.

•  V1 = 3.20 L T1 = 125oC = 398 K •  V2 = 1.54 L T2 = ?

•  T2 = T1 x ( V2 / V1) T2 = 398 K x = 192 K

•  T2 = 192 K oC = K - 273.15 = 192 - 273 oC = -81oC

1.54 L 3.20 L

Avogadro’s Law - Moles and Volume

The volume of gas is directly proportional to the amount of gas (in moles), when measured at the same P and T:

V α n or V = n x constant

n1 = initial moles of gas V1 = initial volume of gas n2 = final moles of gas V2 = final volume of gas

n1 V1

n2 V2

= or: n1 = n2 x V1 V2

Avogadro’s Law: Volume and Amount of Gas

Problem:

Sulfur hexafluoride is a gas used to trace pollutant plumes in the atmosphere, if the volume of 2.67 g of SF6 at 1.143 atm and 28.5 oC is 2.93 m3, what will be the mass of SF6 in a container whose volume is 543.9 m3 at 1.143 atm and 28.5 oC?

Temp and Pressure don’t change

Avogadro's Law Calculate moles of SF6 Convert to moles after

volume change

Convert back to mass of SF6

€

n1V1

=n2V2

Problem: Sulfur hexafluoride is a gas used to trace pollutant plumes in the atmosphere, if the volume of 2.67 g of SF6 at 1.143 atm and 28.5 oC is 2.93 m3, what will be the mass of SF6 in a container whose volume is 543.9 m3 at 1.143 atm and 28.5 oC?

Solution: Molar mass SF6 = 146.07 g/mol

2.67g SF6 146.07g SF6/mol

= 0.0183 mol SF6

V2 V1

n2 = n1 x = 0.0183 mol SF6 x = 3.40 mol SF6 543.9 m3 2.93 m3

mass SF6 = 3.40 mol SF6 x 146.07 g SF6 / mol = 496 g SF6

Avogadro’s Law: Volume and Amount of Gas

IDEAL GAS LAW •  The ideal gas equation combines both Boyles Law and

Charles Law into one easy-to-remember law:

PV=nRT n = number of moles of gas in volume V •  R = Ideal gas constant •  R = 0.08206 L atm / (mol K) = 0.08206 L atm mol-1 K-1

•  An ideal gas is one for which both the volume of molecules and forces between the molecules are so small that they have insignificant effect on its P-V-T behavior.

Variations on Ideal Gas Equation •  During chemical and physical processes, any of the four variables in the

ideal gas equation may be fixed. •  Thus, PV=nRT can be rearranged for the fixed variables:

–  for a fixed amount at constant temperature •  PV = nRT = constant Boyle’s Law

–  for a fixed amount at constant pressure •  V/T = nR/P = constant Charles’ Law

–  for a fixed pressure and temperature •  V = n (RT/P) or V ∝ n Avogadro’s Law

–  for constant volume •  P/T = nR/V = constant when amount constant •  P = nRT/V so P ∝ n when T constant

JUST REMEMBER PV = nRT and rearrange as needed

Standard Temperature and Pressure (STP)

A set of Standard conditions have been chosen to make it easier to quantify gas amounts (i,.e., “liters at STP”).

Standard Temperature = 00 C = 273.15 K

Standard Pressure = 1 atmosphere = 760 mm Mercury

At these “standard” conditions, if you have 1.0 mole of any ideal gas, it will occupy a “standard molar volume”: V = nRT/P =

Standard Molar Volume = 22.414 Liters = 22.4 L or L/mol

Table 5.2 (P 150) Molar Volumes for Various Gases at 00C and 1 atm

Gas Molar Volume (L)

Oxygen (O2) 22.397

Nitrogen (N2) 22.402

Hydrogen (H2) 22.433

Helium (He) 22.434

Argon (Ar) 22.397

Carbon dioxide (CO2) 22.260

Ammonia (NH3) 22.079

Gas Law: Solving for Pressure

Problem: Calculate the pressure in a container whose Volume is 87.5 L and it is filled with 5.038kg of Xenon at a temperature of 18.8 oC. Plan:

Volume is in L

Convert kg of Xe to moles

Convert temp to K

Ideal Gas Law Check value for R

Pressure in atm

Gas Law: Solving for Pressure

Problem: Calculate the pressure in a container whose Volume is 87.5 L and it is filled with 5.038kg of Xenon at a temperature of 18.8 oC.

Solution:

nXe = = 38.37014471 mol Xe 5038 g Xe 131.3 g Xe / mol

T = 18.8 oC + 273.15 K = 291.95 K

PV = nRT so P = nRT V

P = = 10.5 atm (38.37 mol )(0.0821 L atm)(291.95 K) 87.5 L (mol K)

Ammonia Density Problem

•  Calculate the density of ammonia gas (NH3) in grams per liter at 752 mm Hg and 55 oC.

•  P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm •  T = 55 oC + 273 = 328 K PV = nRT so

•  n/V = =

n/V = 0.03674 mol/L density = d = mass per unit volume density = 0.03674 mol/L x (17.03 g/mol NH3)

d = 0.626 g / L (more than 1000x less dense than water!)

P R x T

( 0.989 atm) ( 0.08206 L atm / mol K) ( 328 K)

Ammonia Density Problem – by another method

•  Calculate the density of ammonia gas (NH3) in grams per liter at 752 mm Hg and 55 oC.

Density =d = mass per unit volume = m / L •  P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm •  T = 55 oC + 273 = 328 K n = mass / Molar mass = m / M PV = nRT so m/(MV) = n/V = P/(RT) so d = m/V = PM/(RT)

•  d = =

•  d = 0.626 g / L (more than 1000x less dense than water!)

P x M R x T

( 0.989 atm) ( 17.03 g/mol) ( 0.08206 L atm/mol K) ( 328 K)

Applying the Temperature - Pressure Relationship (constant n and V)

Problem: A copper tank is filled with compressed gas to a pressure of 4.28 atm at a temperature of 0.185 oF. What will be the pressure if the temperature is raised to 95.6 oC? Plan: The volume of the tank is not changed, and we only have to deal with the temperature change, so convert to SI units, and calculate the pressure ratio from the T ratio. Solution:

T1 = (0.185 oF - 32.0 oF)x 5/9 = -17.68 oC T1 = -17.68 oC + 273.15 K = 255.47 K T2 = 95.6 oC + 273.15 K = 368.8 K

P1 P2 T1 T2

= = nR/V

P2 = P1 x = ? T2 T1

P2 = 4.28 atm x = 6.18 atm 368.8 K 255.47 K

Change of Conditions A gas sample in the laboratory has a volume of 45.9L at 25 oC and a pressure of 743 mm Hg. If the temperature is increased to 155 oC by pumping (compressing) the gas to a new volume of 3.10mL, what is the pressure?

Convert all units to K, L, atm Constant moles Set up before and

after equations

Solve for new pressure

Plan:

Change of Condition

= = nR = constant T1 T2

P1 x V1 P2 x V2

( 0.978 atm) ( 45.9 L) P2 (0.00310 L)

( 298 K) ( 428 K) =

P2 = ( 428 K) ( 0.978 atm) ( 45.9 L)

= 2.08 x 104 atm ( 298 K) ( 0.00310 L)

P1= 743 mm Hg x1 atm/ 760 mm Hg = 0.978 atm V1 = 45.9 L V2 = 3.10 ml = 0.00310 L

T1 = 25 oC + 273 = 298 K T2 = 155 oC + 273 = 428 K

IDEAL GAS EQUATION

PV = nRT

Example Problem: Molar Mass of a Gas from its weight and P,V,T

Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 mL flask. If the sample had a mass of 0.118 g at a pressure of 550.0 Torr, what is the molecular weight of the gas? Plan:

Convert all units to K, L, atm

Calculate mass of vapor

Use Ideal Gas Law to calculate n.

Use mole / mass relationship to calculate Molar Mass

Example Problem: Molar Mass of a Gas from its weight and P,V,T

Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 ml flask. If the sample had a mass of 0.118 g at a pressure of 550.0 Torr, what is the molecular weight of the gas? Plan: Use the Ideal gas law to calculate n, then calculate the molar mass. Solution:

P = 550.0 Torr x x = 0.724 atm 1mm Hg 1 Torr

1.00 atm 760 mm Hg

V = 250.0 ml x = 0.250 L 1.00 L 1000 ml

T = 25.0 oC + 273.15 K = 298.2 K n = P V

R T

n = = 0.007393 mol (0.0821 L atm/mol K) (298.2 K)

(0.724 atm)(0.250 L)

M = mass / n = 0.118 g / 0.007393 mol = 16.0 g/mol

Molar Mass of a Gas Problem: A volatile liquid is placed in a flask whose volume is 590.0 ml and allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of 100.0 oC and 736 mm Hg pressure. If the mass of the flask empty and after the experiment was 148.375g and 149.457 g, what is the molar mass of the liquid?

Plan:

Convert all units to K, L, atm

Calculate mass of vapor

Use Ideal Gas Law to calculate n.

Use mole / mass relationship to calculate Molar Mass

Molar Mass of a Gas Problem: A volatile liquid is placed in a flask whose volume is 590.0 ml and allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of 100.0 oC and 736 mm Hg pressure. If the mass of the flask empty was 148.375g and after the experiment 149.457 g, what is the molar mass of the liquid? Plan: Use the gas law to calculate the moles, then its molar mass. Solution:

Pressure = 736 mm Hg x = 0.9684 atm 1 atm 760 mm Hg

mass = 149.457g - 148.375g = 1.082 g

n = (PV/RT) = = 0.0186 mol

M = mass / n = 1.082 g / 0.0186 mol = 58.02 g/mol

(0.0821 L atm/mol K)(373.2 K) ( 0.9684 atm)(0.590 L)

note: the compound is acetone C3H6O = MM = 58g mol !

Chemical Equation Calc - III

Reactants Products Molecules

Moles

Mass Molecular Weight g/mol

Atoms (Molecules) Avogadro’s Number

6.02 x 1023

Solutions

Molarity moles / liter

Gases

PV = nRT

Sodium Azide Decomposition-I

•  Sodium Azide (NaN3) is used in some air bags in automobiles. Calculate the volume of Nitrogen gas generated at 21 oC and 823 mm Hg by the decomposition of 60.0 g of NaN3 .

•  2 NaN3 (s) 2 Na (s) + 3 N2 (g)

•  mol NaN3 = 60.0 g NaN3 / (65.02 g NaN3 / mol NaN3) = = 0.9228 mol NaN3 •  mol N2= 0.9228 mol NaN3 x (3 mol N2/2 mol NaN3) = 1.38 mol N2

Sodium Azide Calc: - II

•  PV = nRT V = nRT/P

•  V =

•  V = 30.8 liters

( 1.38 mol) (0.08206 L atm / mol K) (294 K)

( 823 mm Hg / 760 mmHg / atm )

Figure 5.9: Partial pressure of each gas in a mixture of gases depends on the number of moles of that gas (ni).

Defined as Pi = (ni/ntotal) Ptotal

Ideal Gas Mixtures

•  Ideal gas equation applies to the mixture as a whole and to each gas species (i) individually:

PtotalV = ntotalRT Ptotal is what is measured by

pressure guages. Σ ni = n1+n2+n3…= ntotal

PiV = niRT for all species i. Pi harder to measure.

Note: Σ Pi = P1+P2+P3…= Ptotal

i

i

Partial Pressures

•  A 2.00 L flask contains 3.00 g of CO2 and 0.10 g of Helium at a temperature of 17.0 oC.

•  What are the Partial Pressures of each gas, and the total Pressure?

•  T = 17 oC + 273 = 290 K •  nCO2 = 3.00 g CO2/ 44.01 g CO2 / mol CO2 •  = 0.0682 mol CO2 •  PCO2 = nCO2RT/V •  •  PCO2 =

•  PCO2 = 0.811 atm

( 0.0682 mol CO2) ( 0.08206 L atm/mol K) ( 290 K)

(2.00 L)

Partial Pressures

•  nHe = 0.10 g He / 4.003 g He / mol He •  = 0.025 mol He •  PHe = nHeRT/V

•  PHe =

•  PHe = 0.30 atm

•  PTotal = PCO2 + PHe = 0.812 atm + 0.30 atm •  PTotal = 1.11 atm

(0.025 mol) ( 0.08206 L atm / mol K) ( 290 K )

( 2.00 L )

Mole Fraction of a Species

•  PiV = niRT for all i, and Σ Pi = P1+P2+P3…= Ptotal

• Σ ni = n1+n2+n3…= ntotal

• Σ xi = x1+x2+x3…= 1.00000

•  Pi = niRT/V = (ni/ntotal)ntotalRT/V = xi . Ptotal

•  Partial Pressure of each gas is equal to mole fraction multiplied by the total pressure.

xi = ni/ntotal

Example Problem using mole fractions

•  A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm ?

•  Plan:

Calculate total number of moles

Add up all mole values

Calculate individual mole fractions

Divide individual moles by total moles

Calculate partial pressures

Multiply mole fractions by total pressure

Example Problem using mole fractions

•  A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm ?

•  Total # moles = 4.46 + 0.74 + 2.15 = 7.35 mol = ntotal

•  xNe = nNe / ntotal = 4.46 mol Ne / 7.35 mol = 0.607 •  PNe = xNe PTotal = 0.607 ( 2.00 atm) = 1.21 atm Ne •  xAr = 0.74 mol Ar / 7.35 mol = 0.10 •  PAr = xAr PTotal = 0.10 (2.00 atm) = 0.20 atm Ar •  xXe = 2.15 mol Xe / 7.35 mol = 0.293 •  PXe = xXe PTotal = 0.293 (2.00 atm) = 0.586 atm Xe

. . . .

. . . . .

. . . .

. . . . .

. . .

Saturation

Figure 5.10: Production of O2(g) by thermal decomposition of KCIO3:

It is mixed with vapor pressure of water.

Collection of Hydrogen gas over Water - Vapor pressure - I

•  2 HCl(aq) + Zn(s) ZnCl2 (aq) + H2 (g)

•  Calculate the mass of Hydrogen gas collected over water if 156 mL of gas is collected at 20oC and 769 mm Hg.

•  PTotal = PH2 + PH2O PH2 = PTotal - PH2O FROM TABLE PH2 = 769 mm Hg - 17.5 mm Hg = 752 mm Hg •  T = 20oC + 273 = 293 K •  PH2 = 752 mm Hg /760 mm Hg /1 atm = 0.989 atm •  V = 0.156 L

COLLECTION OVER WATER CONT.-II

•  PH2V = nH2RT nH2 = PH2V / RT

•  nH2 =

•  nH2 = 0.00641 mol •  Mass H2 = 0.00641 mol x 2.01 g H2 / mol H2 = 0.0129 g H2

(0.989 atm) (0.156 L) (0.0821 L atm/mol K) (293 K)

Chemical Equations

Reactants Products Molecules

Moles

Mass Molecular Weight g/mol

Atoms (Molecules) Avogadro’s Number

6.02 x 1023

Solutions

Molarity moles / liter

Gases

PiV = niRT

Gas Law Stoichiometry - I - NH3 + HCl Problem: A slide separating two containers is removed, and the gases are allowed to mix and react. The first container with a volume of 2.79 L contains Ammonia gas at a pressure of 0.776 atm and temperature = 18.7 oC. The second with a volume of 1.16 L contains HCl gas at a pressure of 0.932 atm and a temperature of 18.7 oC. What mass of solid ammonium chloride will be formed, and what will be remaining in the container, and what is the pressure? Plan: This is a limiting reactant problem, so we must calculate the moles of each reactant using the gas law to determine the limiting reagent. Then we can calculate the mass of product, and determine what is left in the combined volume of the container, and the conditions. Solution:

Equation: NH3 (g) + HCl(g) NH4Cl(s)

TNH3 = 18.7 oC + 273.15 = 291.9 K

NH3 HCl

Gas Law Stoichiometry - II - NH3 + HCl

ni = PiV RT

nNH3 = = 0.0903 mol NH3 ÷ 1 (0.776 atm) (2.79 L)

(0.0821 L atm/mol K) (291.9 K)

nHCl = = 0.0451 mol HCl ÷ 1 (0.932 atm) (1.16 L)

(0.0821 L atm/mol K) (291.9 K)

limiting

reactant

Therefore the product will be 0.0451 mol NH4Cl or 2.28 g NH4Cl

Ammonia remaining = 0.0903 mol - 0.0451 mol = 0.0452 mol NH3

V = 1.16 L + 2.79 L = 3.95 L

PNH3= = = 0.274 atm nNH3RT

V (0.0452 mol) (0.0821 L atm/mol K) (291.9 K)

(3.95 L)

Postulates of Kinetic Molecular Theory

The kinetic energy associated with the molecules’ velocity: KE = (1/2) mass x u2

The temperature of matter is a measure of its thermal energy, such that T is proportional to the average KE. so that their velocity (u) is proportional to √T.

Gas molecules make collisions with the walls, and on average simply reverse their momentum in the direction perpendicular to the wall. This gives rise to the pressure (force per unit area) on the wall.

Figure 5.11: An ideal gas particle in a cube whose sides are of length L

(in meters).

Figure 5.12: (b) The velocity u of any gas particle can be broken down into three mutually perpendicular components, ux, uy, and uz.

Momentum = mass x velocity = m.u

Figure 5.13: (a) In collision: exact reversal of the x component of the velocity, so that momentum change on impact is 2 x mass x ux. (b) The # impacts on the shaded walls per unit time can be calculated from the velocity and the number density. Proportional to both. (c) Force on wall = the average momentum change per unit time =

(momentum change per impact) x (# impacts per unit time)

Velocity and Energy •  Kinetic Energy = KE = ½ mu2 for one molecule •  Average Kinetic Energy of a mole of gas (KEavg) = 3/2 RT independent of gas identity (use R in J/mol K) •  Average Kinetic Energy of one molecule (KEavg)

= 3/2 RT / NA independent of mass, identity = ½ mu2 (mean value of u2)

½ mu2 = 3/2 RT/NA u2 = 3RT/(mNA) = 3RT/M since m = M/NA

√(u2) = √(3RT/M) “Root mean square velocity”

Must use R in J/mol K, and remember that 1 J = 1 kg m2/s2

Figure 5.16: A plot of the relative number of N2 molecules that have a given velocity at three temperatures.

Also: At same T, heavier gases go slower, as 1/√M: Average u2/u1 = (M1/M2)1/2

Kinetic energy per mole = ½ Mu2 = 3/2 RT, independent of M HEAT!

Velocity and Energy •  Kinetic Energy = KE = ½ mu2 for one molecule •  Average Kinetic Energy of a mole of gas (KEavg) = 3/2 RT independent of gas identity •  Average Kinetic Energy of one molecule (KEavg)

= 3/2 RT/NA independent of mass, identity = ½ mu2 (mean value of u2)

u2 = 3RT/(mNA) = 3RT/M

√(u2) = √(3RT/M) “Root mean square velocity”

Must use R in J/mol K, and remember that 1 J = 1 kg m2/s2

Molecular Mass and Molecular Speeds - I

Problem: Calculate the molecular speeds of the molecules of hydrogen, methane, and carbon dioxide at 300K! Plan: Use the equations for the average kinetic energy of a molecule, and the relationship between kinetic energy and velocity to calculate the average speeds of the three molecules. Solution: for Hydrogen, H2 = 2.016 g/mol

EK = x T = 1.5 x x 300K = 3 R 2 NA

8.314 J/mol K 6.022 x 1023 molecules/mol

EK = 6.213x10 - 21 J/molecule = 1/2 mu2

m = = 3.348x10 - 27 kg/molecule 2.016 x 10 -3 kg/mole 6.022 x 1023 molecules/mole

EK = 6.213x10 - 21 J/molecule = 1/2 mu2 = 1.674x10 - 27 kg/molecule (u2)

u = 1.926 x 103 (J/kg)1/2 = 1.926 x 103 m/s = 1,926 m/s

Molecular Mass and Molecular Speeds - II

For methane , CH4: M = 16.04 g/mole = 16.04 x 10 - 3 kg/mole

√(u2) = √(3RT/M) R = 8.314 J/mol K = 8.314 kg m2/(s2 mol K)

√(u2) = √ [(3x 8.314 kg m2/(s2 mol K) x 300 K)/ 16.04 x 10 - 3 kg/mole] Note: 1 J = 1 kg m2/s2

√(u2) = 6.838 x 102 m/s = 683.8 m/s Root mean square velocity

Molecular Mass and Molecular Speeds - III

for Carbon dioxide CO2 = 44.01 g/mole

u = 4.124 x 102 (J/kg)1/2 = 412.4 m/s

6.213 x 10 - 21 J/molecule = 3.654 x 10 - 26 kg/molecule (u2)

m = = 7.308 x 10 - 26 kg/molecule 44.01 x 10 - 3 kg/mole 6.022 x 1023 molecules/mole

EK = 6.213 x 10 - 21 J/molecule = 1/2mu2

EK = x T = 1.5 x x 300K = 3 R 2 NA

8.314 J/mol K 6.022 x 1023 molecules/mole

Molecular Mass and Molecular Speeds - IV

Molecule H2 CH4 CO2

Molecular Mass (g/mol)

Kinetic Energy (J/molecule)

Velocity (m/s) ∝ 1/√M

2.016 16.04 44.01

6.213 x 10 - 21 6.213 x 10 - 21 6.213 x 10 - 21

1,926 683.8 412.4

Diffusion vs Effusion (Rates go as gas velocity)

•  Diffusion - One gas mixing into another gas, or gases, of which the molecules are colliding with each other, and exchanging energy between molecules.

•  Effusion - A gas escaping from a container into a vacuum (where there are very few molecules to block it’s molecules’ paths)

Diffusion

•  Grahams Law – Rate of diffusion is inversely proportional to

the square root of the mass of the gas. •  For two gases rate of diffusion is:

€

=M2

M1

Rate of diffusion Gas 1 Rate of diffusion Gas 2

Figure 5.19: (a) demonstration of the relative diffusion rates of NH3 and HCL

molecules through air.

Figure 5.19: (b) When HCl(g) and NH3 (g) meet in the tube, a white ring of

NH4Cl(s) forms.

Diffusion Rates: proportional to average velocity or to 1/√M

•  Rate1/Rate2 = u1/u2 = (M2/M1)1/2

•  HCl = 36.46 g/mol NH3 = 17.03 g/mol

•  RateNH3 = RateHCl x ( 36.46 / 17.03 )1/ 2 •  RateNH3 = RateHCl x 1.463

Figure5.18:Effusionrateofagasintoanevacuatedchamber=collisionratewithimaginarysurface.